Question 374269
If y varies as x cubed, then we have : y(x)=x^3 
 
let double y, and find the new value of x called x' : y'(x) = 2y(x) = x'^3, 
 
x' = (2y(x))^(1/3) = 2^(1/3)(x^3)^(1/3) = 2^(1/3)x, 
  
 hence x has to be multiplied by the 3rd-root of two, or by 2^(1/3)